Question

Evaluate the integral...

Answer 1

Not sure what to do with the x^-3 on this one.

Answer 2

you can "distribute the integration..
\[\int(x^{-3}-8)dx=\int x^{-3}dx-8\int dx\]

Answer 3

The antiderrivative should be (((x^-2)/-2) -8x) evaluated from 1/2 to 1.

Answer 4

now you can apply the resular rules..

Answer 5

Just apply the inverse power rule to both

Answer 6

Cool! I get it it now.

Answer 7

\[\int\limits_{}^{}x ^{-3}-8=x ^{-2}/-3-8x\]

Answer 8

@sd_ghafouri your integral rule is wrong...
the denominator is
-2
not -3
\[\int(x^{-3}-8)dx={x^{-2}\over-2}-8x\]

Answer 9

and o'course you are asked to use the fundamental rule for definite integration... so, apply the limits

Answer 10

I evalute the true one just problem with typing sarry

Answer 11

yours is rigth

Answer 12

okay I'm getting\[[-\frac{ 1 }{ 2 }(1^{-2})-8]-[-\frac{ 1 }{ 2 }(\frac{ 1 }{ 2 }^{-2})-8 = \frac{ 3 }{ 2 }\]

Answer 13

Nevermind. Forgot to multiply 8 * 1/2. -5/2